A) 6.56 ´ 10?8 eV
B) 2 ´ 10?8 eV
C) 10?8 eV
D) 8 ´ 10?8 eV
Correct Answer: A
Solution :
The average time that the atom spends in this excited state is equal to Dt, so by using \[\Delta E.\,\Delta t=\frac{h}{2\pi }\] Þ Uncertainty in energy\[=\frac{h/2\pi }{\Delta t}\] \[=\frac{6.6\times {{10}^{-34}}}{2\times 3.14\times {{10}^{-8}}}=1.05\times {{10}^{-26}}J=6.56\times {{10}^{-8}}eV\]
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